Saturday, December 10, 2005

The lies of a salesman

To uniformly transfer the heat, the cookware is made from 5-layers of metals sandwiched together. Outer layers are T304 stainless steel, then 1145 aluminum, and the center layer is 3004 aluminum-alloy.

That was lie number 2.

Since thermal conductivies of T304 stainless steel (16.2 W/m-K), 1145 aluminum (230 W/m-K), and 3004 aluminum-alloy (163 W/m-K) are different, sandwich them together would not improve thermal conductivity, rather it will create thermal gradients.

It gets even better:

One other claim Classica has on their website is that their cookware will decrease energy usage. Also, some readers have commented that cookware with layered construction does improve uniform surface heating.

Since I personally do not own any cookware that is constructed in such way, I can only rely to a textbook called “Heat Transfer, A Practical Approach” by Yunus A. Cengel, ISBN 0-07-011505-2.

For this calculation, a positive Q value is the energy used to increase maintain a set temperature; T1 and T2 are the temperatures at inner and outer surfaces of the cookware, R is the cookware’s thermal resistivity, or conduction resistance, L is the cookware’s thickness, k is thermal conductivity, and A is the cookware heating surface area.

Assume we have two pans; both are 12” (30.48 cm) in diameter and 5 mm thick. One is made from the 5-layer construction described via Classica cookware’s website, with each layer 1 mm thick, and the other is a plain T304 stainless steel.

Let T1-T2 equal 10 C. In other words, we are looking for how much energy is needed to maintain 10 C in the pan.

From my own calculation, approximately 2,400 W of energy is needed to maintain 10 C in the plain stainless steel pan. Close to 5,300 W of energy is needed to achieve the same result in the 5-layer pan by Classica cookware.

Therefore, to maintain same temperature, Classica’s cookware requires more than double the energy than an ordinary stainless steel pan.